To make the answer complete I first repeat some of my comment.
Suppose $J$ is the Jordan's canonical form of $A$, with $J=PAP^{-1}$.
Then we have
\begin{align*} &A^4 + 2A^3 + A^2 = 0 &\\ \iff & P(A^4 + 2A^3 + A^2 )P^{-1}= 0 & (\text{since $P0P^{-1}=0$})\\ \iff & J^4 + 2J^3 + J^2 = 0 & (\text{since $PA^nP^{-1}=PAP^{-1}PAP^{-1}\cdots PAP^{-1}=J^n$}) \end{align*}
The same argument gives $J^2+J\ne 0$.
So it suffices to determine all the Jordan-canonical form $J$ s.t. $J^4 + 2J^3 + J^2 = 0$, $J^2+J\ne 0$.
There are three cases according to the sizes and numbers Jordan blocks. We are going to check them case by case.
For convinience denote $f(x)=x^4+2x^3+x^2$, $ g(x)=x^2+x $. Note that $f(x_0)=0$ $\implies$ $x_0\in\{0,-1\}$ $\implies$ $g(x_0)=0$.
Case 1: $J$ has Jordan blocks "$1+1+1$". (That is it has three Jordan blocks, each with size one.)
Then $J=\mathtt{diag}\{\lambda,\mu, \omega\}$, where $\lambda,\mu, \omega$ need not be distinct.
Since $J^4 + 2J^3 + J^2 = 0$, we have $f(\lambda)=f(\mu)=f(\omega)=0$.
Therefore $\lambda,\mu, \omega\in\{0,-1\}$.
But this implies $\lambda,\mu, \omega$ are zeros of $g(x)$.
Hence $J^2+J=\mathtt{diag}\{g(\lambda),g(\mu), g(\omega)\}=0$, a contradiction.
Therefore Case 1 is impossible.
Case 2: $J$ has Jordan blocks "$2+1$".
Then
\begin{equation*}
J=\begin{pmatrix} \lambda & 1 & \\ & \lambda & \\ & & \mu \end{pmatrix},
\end{equation*}
where $\lambda,\mu$ need not be distinct.
A direct computation gives
\begin{equation*}
J^4 + 2J^3 + J^2 =\begin{pmatrix}
f(\lambda) & 4\lambda^3+6\lambda^3+2\lambda^2 & \\
& f(\lambda) & \\
& & f(\mu)
\end{pmatrix}.
\end{equation*}
So we must have $f(\lambda)=f(\mu)=4\lambda^3+6\lambda^3+2\lambda^2=0$.
This gives $\lambda,\mu \in\{-1,0\}$.
Another direct computation gives
\begin{equation*}
J^2+J=\begin{pmatrix}
\lambda^2+\lambda & 2\lambda+1 & \\
& \lambda^2+\lambda & \\
& & \mu^2+\mu
\end{pmatrix}.
\end{equation*}
When $\lambda\in\{0,-1\}$, $2\lambda+1\ne 0$.
Therefore $J^2+J\ne 0$ whenever $\lambda\in\{0,-1\}$.
Hence case 2 gives four solutions:
\begin{equation*}
J=\begin{pmatrix} \lambda & 1 & \\ & \lambda & \\ & & \mu \end{pmatrix},
\end{equation*}
where $\lambda,\mu\in\{0,-1\}$.
Case three: $J$ has Jordan blocks "$3$".
Then
\begin{equation*}
J=\begin{pmatrix}
\lambda & 1 & \\
& \lambda & 1 \\
& & \lambda
\end{pmatrix} .
\end{equation*}
A direct proof gives
\begin{equation*}
J^4 + 2J^3 + J^2=\begin{pmatrix}
f(\lambda) & \text{the value here is not important} & 4\lambda^2+3\lambda+1 \\
& f(\lambda) & \text{the value here is not important} \\
& & f(\lambda)
\end{pmatrix} .
\end{equation*}
Since $f(\lambda)$ and $ 4\lambda^2+3\lambda+1 $ can not be zero at the same time, we conclude Case three is impossible.
To sum up there are (up to similarity) four choices for $J$ (and hence for $A$):
\begin{equation*}
J=\begin{pmatrix} \lambda & 1 & \\ & \lambda & \\ & & \mu \end{pmatrix},
\end{equation*}
where $\lambda,\mu\in\{0,-1\}$.
